I'm having trouble with a seemingly simple statement.

Suppose $N|K$ is a Galois extension with $L$ an additional field over $K$. Now for $\sigma \in \text{Gal}([N,L] | L)$, with $[N,L]$ being the composite of $N$ and $L$, I'd like to show that $\sigma(N) \subset N$.

I'm aware of the implication from the Fundamental Theorem of Galois Theory that says that for $K\leq E\leq F$ with $F|K$ and $E|K$ Galois extensions, $\sigma \in \text{Gal}(F|K): \sigma(E) \subset E$, which would solve my problem. But I can't seem to show that $[N,L]|K$ is Galois (I know an extension being Galois is not transitive). Does it suffice to say that $[N,L]$ is also the splitting field over $K$ of a separable polynomial, just like $N$, although $[N,L]$ would eventually contain more elements?

This is my first question here, so let me know if I can improve the format of my question.

  • $\begingroup$ Let $M/K$ be a finite field extension, and $E,F$ two intermediate fields $E,F$. If $E/K$ is Galois, then $EF/F$ is Galois. If $E/K$ and $F/K$ are both Galois, then $EF/K$ is Galois. $\endgroup$ Feb 24, 2023 at 9:57
  • $\begingroup$ Thank you. Both points are clear to me. So that means, generally, $EF|K$ is not Galois, if for example $F|K$ is not Galois. Do you have an idea how to show the statement in the first sentence of my question without having $EF|K$ be a Galois extension? $\endgroup$
    – SpinnyCube
    Feb 24, 2023 at 10:17
  • $\begingroup$ Just to give an easy example of $EF/K$ not generally being Galois: Take a non-Galois extension $L/K$ with $N\subset L$ (you know such examples, because you know that being Galois is not transitive). Then the compositum $NL=L$, so $NL/K$ is not Galois. $\endgroup$
    – CPCH
    Feb 24, 2023 at 15:48

1 Answer 1


Let $M/K$ be a finite field extension, and $E,F$ two intermediate fields with $E/K$ Galois. Then $EF/F$ is Galois and there is an injective group homomorphism $$ \mathrm{Gal}(EF/F) \to \mathrm{Gal}(E/K), \quad \sigma \mapsto \sigma|_E $$ To see this, write $E$ as the splitting field of a separable polynomial $f\in K[X]$. Then $E/K$ is generated by the roots of $f$, so $EF=F(E)$ is generated over $F$ by the roots of $f$. Thus $EF/F$ is the splitting field of $f\in F[X]$, which is still separable, and hence $EF/F$ is Galois.

Now, if $\sigma\in\mathrm{Gal}(EF/F)$, then $\sigma$ permutes the roots of $f$, so restricts to an automorphism of $E$. It is easy to check that we have a group homomorphism between the Galois groups. Finally, if $\sigma$ is the identity on $E$, then it fixes all the roots of $f$, so $\sigma$ is the identity on $EF$, and the group homomorphism is injective.

  • $\begingroup$ Thank you! Thinking of an element of $\text{Gal}(EF|F)$ as a permutation on the roots of the same separable polynomial as the one whose splitting field over $K$ is $E$ was very helpful! $\endgroup$
    – SpinnyCube
    Feb 24, 2023 at 17:10

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